The least value of the product $$xyz$$ (such that $$x$$, $$y$$ and $$z$$ are positive real numbers) for which the determinant $$\begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix}$$ is non-negative is

$$\Delta = x(yz - 1) - (z - 1) + (1 - y)$$

$$\Delta = xyz - (x + y + z) + 2$$

$$xyz - (x + y + z) + 2 \ge 0$$

$$\Rightarrow xyz + 2 \ge x + y + z$$

$$\frac{x + y + z}{3} \ge \sqrt[3]{xyz}$$

$$\Rightarrow x + y + z \ge 3(xyz)^{1/3}$$

$$xyz + 2 \ge 3(xyz)^{1/3}$$

$$k^3 + 2 \ge 3k$$

$$(k - 1)^2(k + 2) \ge 0$$

Since $$(k-1)^2$$ is always non-negative, the inequality holds as long as $$k + 2 \ge 0$$, which means $$k \ge -2$$.

If $$k \ge -2$$, then $$(xyz)^{1/3} \ge -2$$. Cubing both sides, $$xyz \ge (-2)^3 = -8$$.